Single Variable Calculus, Chapter 2, 2.4, Section 2.4, Problem 13

(a) Determine a number $\delta$ such that if $| x - 2 | < \delta$ then $| 4x - 8 | < \varepsilon$, where $\varepsilon = 0.1$

Based from the definition,

$
\begin{equation}
\begin{aligned}

\begin{array}{c}
\text{ if } |x -a| < \delta \text{ then } |f(x) - L| < \varepsilon\\
\text{ if } |x -2| < \delta \text{ then } |4x - 8| < 0.1

\end{array}

\end{aligned}
\end{equation}
$


To satisfy inequatlity $| x - 2 | < \delta $

We want,

$
\begin{equation}
\begin{aligned}
& | 4x - 8 | < 0 . 1
&& \phantom{x}\\

& | 4(x - 2) | < 0.1
&& \text{ Factor}\\

& \frac{4 | x - 2 |}{4} < \frac{0.1}{4}
&& \text{ Divide both sides by 4}\\

& | x- 2| < 0.025
&&

\end{aligned}
\end{equation}
$


Hence,
$\quad \delta < 0.025$

(b) Repeat part (a), where $\varepsilon = 0.01$

Using the definition,

$
\begin{equation}
\begin{aligned}

& | 4x - 8 | < 0.01
&& \\
& 4| x - 2 | < 0.001
&& \text{ Factor }\\

& \frac{4| x-2|}{4} < \frac{0.01}{4}
&& \text{ Divide both sides by 4}\\

& |x-2| < 0.0025
&& \\


\end{aligned}
\end{equation}
$

Hence,
$\quad \delta < 0.0025$

This means that by keeping $x$ within $0.0025$ of $2$, we are able to keep $f(x)$ within $0.1$ of $8$.

Although we chose $\delta = 0.0025$, any smaller positive value of $\delta$ would also have work.